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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS2 Marks
Question
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$\text{y}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2$ : $\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\frac{\text{dy}}{\text{dx}}-\text{xy}+\text{x}^2-2=0$
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Answer

$\text{Here,}\ \ \text{y}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2$
$\therefore\ \ \frac{\text{dy}}{\text{dx}}=\text{a e}^{\text{x}}-\text{b e}^{-\text{x}}+2\text{x}$
$\therefore\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+2$
$\therefore\ \ \text{L.H.S.}=\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}+\text{x}^2-2$
$=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+2-(\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2)+\text{x}^2-2$
$=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+2-\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2+\text{x}^2-2$
$=0=\text{R.H.S.}$
$\therefore\ \ \text{y}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2$
is a solution of the given differential equation.

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